Question: Let $g(x)=\dfrac{\cos(x)}{\sin(x)}$. Find $g'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{1}{\sin^2(x)}$ (Choice B) B $\dfrac{1}{\sin^2(x)}$ (Choice C) C $\dfrac{\sin^2(x)-\cos^2(x)}{\sin^2(x)}$ (Choice D) D $\dfrac{\cos^2(x)-\sin^2(x)}{\sin^2(x)}$
Answer: $g(x)$ is the quotient of two, more basic, expressions: $\cos(x)$ and $\sin(x)$. Therefore, the derivative of $g$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left[\dfrac{\cos(x)}{\sin(x)}\right] \\\\ &=\dfrac{\dfrac{d}{dx}[\cos(x)]\sin(x)-\cos(x)\dfrac{d}{dx}[\sin(x)]}{[\sin(x)]^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{-\sin(x)\cdot\sin(x)-\cos(x)\cdot\cos(x)}{\sin^2(x)}&&\gray{\text{Differentiate }\cos(x)\text{ and }\sin(x)} \\\\ &=\dfrac{-[\sin^2(x)+\cos^2(x)]}{\sin^2(x)}&&\gray{\text{Simplify}} \\\\ &=-\dfrac{1}{\sin^2(x)}&&\gray{\text{Pythagorean identity}} \end{aligned}$ Note: Since $\dfrac{\cos(x)}{\sin(x)}=\cot(x)$, we have just shown that $\dfrac{d}{dx}[\cot(x)]=-\dfrac{1}{\sin^2(x)}$. In conclusion, $g'(x)=-\dfrac{1}{\sin^2(x)}$